(ii) It was found that the mean length of 200 diodes (LED) produced by a company was 20.04 mm with a standard deviation of 0.02mm. Find the probability that a diode
selected at random would have a length less than 20.01mm

6.68% probability that a diode selected at random would have a length less than 20.01mm

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Find the probability that a diode selected at random would have a length less than 20.01mm

This is the pvalue of Z when X = 20.01. So

has a pvalue of 0.0668

6.68% probability that a diode selected at random would have a length less than 20.01mm

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

answer

false

explanation

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